There exist two complex numbers c, say c1 and c2, so that 3+2i, 6+i, and c form the vertices of an equilateral triangle. Find the product c1c2 in rectangular form.

littlemixfan Aug 29, 2020

#1**+2 **

I'd just do it like a co-ordinate geometry question

(3,2), (6,1)

\(distance=\sqrt{9+1}=\sqrt{10}\)

height from midpoint =\(\sqrt{10+2.5}=\sqrt{12.5}\)

le

gradient = 1/-3 gradient of perpendicular = 3

midpoint = (4.5, 1.5)

Equ of perpendicular

Now I would find the equation of the perpendicular bisector

And I would find the equation of the circle centre (4.5,1,5) radius sqrt12.5

and solve them simultaneously.

This is not a very 'profession' method since they are complex numbers, not co-ordinate geometry, but the method will work.

\(\)

Melody Aug 29, 2020